Rational maps with all critical points fixed

Consider $$z \mapsto \frac{(n-2) z^n + n z}{n z^{n-1} + (n-2)}.$$ This has $n+1$ fixed points, at $0$, $\infty$, and the $(n-1)$-st roots of $-1$. The only critical points are the roots of $-1$, each of which is ramified of index $3$. So this is a map with all critical points fixed, and all fixed points but two critical.

I am tempted to leave it at that. But, being a nice person, I will explain how I found this. Moreover, I will show that this is (up to conjugacy), the only degree $n$ map with $n+1$ distinct fixed points, two of which are not critical and the rest of which are critical with multiplicity $2$ (ramification index $3$.)

We can take the two noncritical fixed points to be $0$ and $\infty$. So our map is of the form $$z \mapsto z - z p/q,$$ where $p$ and $q$ are of degree $n-1$ and relatively prime. The $n-1$ fixed points other than $0$ and $\infty$ are the roots of $p$.

The derivative of this map is $$\frac{q^2 - zqp' + zpq' - pq}{q^2}.$$ The condition that the fixed points other than $0$ and $\infty$ be critical means $p$ divides the numerator. So $p$ divides $q (q-zp')$ and, as $p$ and $q$ are relatively prime, we conclude that $q - z p' = kp$. Checking degrees, $k$ has degree $0$ and is thus a constant, to be determined later.

Now, we want to impose the stronger condition that every zero of $p$ be doubly a critical point, so the numerator is $\ell p^2$ for some constant $\ell$. Plugging in $q = kp + z p'$, and simplifying $$\ell p^2 = p \left( k(k-1) p + 2 k z p' + z^2 p'' \right).$$ Cancelling $p$ from both sides, $$\ell p = k(k-1) p + 2 k z p' + z^2 p''.$$ Plugging in $z=0$, and noting that $p(0) \neq 0$, we get $\ell = k(k-1)$. So $$2 kz p' + z^2 p'' =0.$$ The solution to this differential equation is $p = C z^{1-2k} + D$. But we know that $z$ has degree $n-1$, so $1-2k=n-1$ and $k = -(n-2)/2$.

Taking the simplest choices $C=D=1$ and plugging back in gives the above solution. All other solutions are related to this one by rescaling the variable $z$.


I was thinking about this very old question again, due to conversations with Sarah Koch. Here is another infinite family of examples which we can write down explicitly: Let $g(x)$ be the degree $n$ Gegenbauer polynomial satisfying $$g'' = \frac{n(n-1)}{x^2-1} g.$$ Let $f(x)$ be the Newton recursion $$f(x) = x-g/g'.$$ Then $$f' = \frac{g g''}{(g')^2} = \frac{n (n-1) g^2}{(x^2-1) (g')^2} .$$ So every critical point of $f$ is a root of $g$, and hence a fixed point of $f$.

More specifically, $\pm 1$ are critical points of multiplicity $1$ and the other $n-2$ roots of $g$ have multiplicity $2$. The only fixed point which is not critical is the fixed point at $\infty$.

On reading further, I see that this example is in section 11 of the Cordwell et al paper.


There has been a lot of research on this question in the last few years. As well as the paper of Cordwell et al linked by the OP, see the Ph. D thesis of Hlushchanka and this paper of his.


David Speyer's answer is right on the money. Douady and Hubbard proved that a map of the sphere to itself whose postcritical set is finite has a unique "uniformization" as a rational map (up to conjugation by Mobius transformations). For a given degree, then, there are finitely many rational maps with all critical points fixed points, since these are postcritically finite.

I haven't thought about the second part of the question about whether one can bound the degree if there are only two non-critical fixed points. It might be possible to determine this from the branching data and the Lefschetz fixed-point formula.


I've been thinking about this a little. I would guess that, for any sufficiently large $n$, there is a finite, nonzero, number of rational maps of degree $n$ such that all of the critical points are fixed. Here is my heuristic argument.

Fix a partition of $2n-2$ into $n+1$ parts: $2n-2 = \lambda\_1 + \lambda\_2 + \cdots \lambda_{n+1}$. For any $n+1$ points $z\_1$, $z\_2$, ..., $z\_{n+1}$ on $\mathbb{CP}^1$, there are finitely many degree $n$ covers of $\mathbb{CP}^1$ which are ramified over the $z\_i$, with the ramified point over $z\_i$ being ramified of index $\lambda\_i+1$, and no other ramification. (You just need to choose which $\lambda\_i$ sheets will be permuted by the monodromy around $z\_i$.) Some of these covers will be disconnected, but all the connected ones will have genus $0$ by the Riemmann-Hurwitz formula. FIRST NONRIGOROUS STEP: I expect that, for most choices of the $\lambda\_i$, there will be a nonzero number of connected covers. Let D be the number of these connected covers.

Now, in each of these connected covers, the covering curve has genus $0$, and is thus isomorphic to $\mathbb{CP}^1$. Let $w\_i$ be the ramified preimage of $z\_i$. The $n+1$ points $w\_i$ give us a point in $M\_{0,n+1}$, and the points $z\_i$ give another point of $M_{0,n+1}$. Plotting the pairs $((w\_1, w\_2, \ldots, w\_n), (z\_1, z\_2, \ldots, z\_n))$ gives us a subvariety of $M\_{0, n+1} \times M\_{0,n+1}$ of dimension equal to that of $M\_{0,n+1}$; the projection onto the second factor is generically $D$ to $1$. Let's call this subvariety $X$. You goal is to understand the intersection of $X$ with the diagonal.

Now, here is the VERY NONRIGOROUS STEP. $X$ has dimension $(n+1)-3=n-2$. So does the diagonal. Our ambient space, $M\_{0, n+1} \times M\_{0,n+1}$, has dimension $2n-4$. In the absence of any other information, the intersection is probably finite and nonempty. :-)

I expect we may be able to extend all of these ideas to work with subvarieties of the compactification $\overline{M}\_{0,n+1}$. That would be good because then we could hope to compute the cohomology class of $X$, and show that it cannot miss the diagonal.

Filling in the gaps here sounds like a really nice problem. Unfortunately, I have too many nice problems, but I wish you luck.